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I SR pseudo-paradox: photons affected by emitter

  1. Jun 11, 2016 #1
    This question is a new (to me) wrinkle on the old Special Relativity spaceship pseudo-paradox gedankens.

    Suppose you observe two spaceships motionless relative to you, side-by-side a mile apart.

    First, a rifle is fired from one at a target on the other, the bullet hits the target. Now they speed up to 2000 mph. The rifle is fired again, hits the target. There are two ways to look at this.

    In spaceships' inertial frame they're still holding still relative to one another. By Galilean relativity, naturally the bullet hits the target again.

    From observer's point of view the rifle, and the bullet, have acquired momentum from their spaceship. The bullet therefore doesn't leave the spaceship at 90 degrees as before. Instead (suppose the bullet speed is inherently 2000 mph) it goes at a 45 degree angle ahead, at 2828 mph, and hits the target.

    Ok, now use a laser instead of a rifle. And, the spaceships speed up to .99 c. Everything else unchanged.

    In spaceships' inertial frame they're still holding still relative to one another. By special relativity, naturally the photons from the laser hit the target again.

    From observer's point of view, since the target is still hit by the photons, the path they take appears angled ahead about 45 degrees. Of course speed is still c, so it appears to take longer to cross the intervening space, but that's immaterial.

    The question is: can we say the photons emitted transversely from the laser on the speeding spaceship acquired momentum from it, just like the rifle bullet?

    Full disclosure: this is a trick question!

    This quote is from "On The Electrodynamics of Moving Bodies", A. Einstein, translated by W. Perrett and G.B. Jeffery from original paper: "Zur Elektrodynamik bewegter Korper", Annalen der Physik, 17, 1905. First published by Methuen and Company, Ltd, 1923. Wikipedia, BTW, has the same quote.

    The second postulate: "... light is always propagated in empty space with a definite velocity c which is independent of the state of motion of the emitting body."

    Note that it says the "velocity" is independent of emitter motion, not "speed".
     
    Last edited: Jun 11, 2016
  2. jcsd
  3. Jun 11, 2016 #2

    PeterDonis

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    I'm not sure what your point is. The photon's velocity is independent of the emitter's state of motion, but the photon's momentum is not, because, for massless particles, the momentum is not a function of velocity. So the photon can still gain momentum from the emitter, just like the rifle bullet does. So what's the problem?
     
  4. Jun 11, 2016 #3

    Dale

    Staff: Mentor

    You are reading too much into this. This is an English translation of German writing. Ordinary language is inherently imprecise, especially when translated. The math is precise and makes the meaning clear.

    The math is the same for both the bullet and the laser. All relativistic laws use the relativistic velocity addition formula. Simply apply that to both the bullet and the light and you get the correct answer in each case.
     
  5. Jun 11, 2016 #4

    jtbell

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    I could be mistaken about this, but I seem to remember reading that the distinction between speed as a scalar and velocity as a vector didn't become common in English-language textbooks until after Einstein's famous papers, and that even today it doesn't apply as strongly (maybe not at all) in languages other than English.

    [added] The Wikipedia article about velocity notes that "the likely origin of the speed/velocity terminology in vector physics" is a textbook on vector analysis by E. B. Wilson, published in 1901.
     
  6. Jun 11, 2016 #5
    That's got to be it, alright, the word "velocity" just means "speed" here. Someone on another board thought he'd found an "Einstein mistake", but it's just terminology. I should have checked that Wikipedia article. Well, it's a good thing to be aware of. Thanks jtbell et al.
     
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